Question

Test the claim that the proportion of people who own cats is larger than 70% at the 0.10 significance level.

Answer 1

Answer:

H0:p≤0.7

H1:p>0.7

right tailed test

$z=\frac{0.75 -0.7}{\sqrt{\frac{0.7(1-0.7)}{400}}}=2.182$  

$p_v =P(z>2.182)=0.0146$  

Since the p value is lower than the significance level given of 0.1 we have enough evidence to reject the null hypothesis.

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test if the  the proportion of people who own cats is larger than 70% at the 0.10 significance level so then the best system of hypothesis are:

H0:p≤0.7

H1:p>0.7

And for this case if we analyze the alternative hypothesis we see that we are conducting a right tailed test

Data given

n=400 represent the random sample taken

$\hat p=0.75$ estimated proportion for the people with cats

$p_o=0.7$ is the value that we want to test

represent the significance level

z would represent the statistic

$p_v$ represent the p value

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the data we got:

$z=\frac{0.75 -0.7}{\sqrt{\frac{0.7(1-0.7)}{400}}}=2.182$  

The p valye for this case would be:

$p_v =P(z>2.182)=0.0146$  

Since the p value is lower than the significance level given of 0.1 we have enough evidence to reject the null hypothesis.

Reject the null hypothesis